Question

What is the speed of a projectile at a point on its path where it makes an angle of ${45}^{\circ }$ with the horizontal? It is given that the radius of curvature ($R$) of particle at this point is $50\sqrt{2}\text{m}$. Take $g=10{\text{m/s}}^2$.

• A. $250\sqrt{5}\text{m/s}$
• B. $10\sqrt{2}\text{m/s}$
• C. $10\sqrt{5}\text{m/s}$
• D. $250\sqrt{2}\text{m/s}$

Answer 1

The correct option is C $10\sqrt{5}\text{m/s}$

For finding the radius of curvature we apply the concept of centripetal acceleration, where we resolve the component of acceleration ($g$) of particle towards the centre of circle with radius of curvature ($R$) as radius.


$a_c=\frac{v^2}{r}$..........($1$)
From Eq.($1$) replacing $a_c\to g\cos \theta ,r\to R$:

$R=\frac{v^2}{g\cos \theta }$
$\Rightarrow v^2=R\times g\cos \theta$
$\Rightarrow v^2=50\sqrt{2}\times 10\cos {45}^{\circ }$
$\therefore v=10\sqrt{5}\text{m/s}$